The percentage purity of commercial #"HCl"# is 35% (w/w) & the specific gravity of the solution is 1.18 #"g"//"mL"#. What are the molarity, molality and normality?
1 Answer
I happen to know (from using the following) that
It's dissolved in water, and
#35.453+1.0079 = "36.4609 g/mol"# .
#"1.18 g"/cancel"mL"cdot (1000cancel("mL"))/cancel("1 L")cdotcancel"1 L" = "1180 g"# .
Thus,
Molarity is
#"Molarity" = [413cancel("g")*("1 mol")/(36.4609cancel("g"))]/("1 L") ~~ color(blue)("11.3272 M")#
which is sensible considering how
Molality is
The density of water at
#"Molality" = [413cancel("g")*("1 mol")/(36.4609cancel("g"))]/("0.9970749 kg") ~~ color(blue)("11.3604 m")#
Since
#"Normality"# #~~# #color(blue)("11.3272 N")#